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I've just started with ring theory and want to understand what in actual does it mean by generators of a ring.

As in group theory we have the set of generators as those elements on which if we repeatedly apply binary operation of the group we get all elements of group.But in rings we have two binary operations, so do we choose generators in a ring such that it is generator of that $R$ under addition and also satisfies the associative law under multiplication and distributive law...Can anyone please explain using example? also how do we formally represent generators of a ring ...

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If $R$ is generated by the set $S=\{s_1,\dots, s_n\}$, it means that every element of $R$ can be written in the form $$ \pm(s_{i_{1,1}}s_{i_{1,2}}\dots s_{i_{2,1}})\pm(s_{i_{2,2}}\dots s_{i_{2,p_1}})\pm\dots\pm(s_{i_{n,1}}\dots s_{i_{n,p_n}}) $$ In other words, each element of $R$ can be written as a sum or difference of elements, each of which is a product of elements in $S$. For example, consider the ring of polynomials in $x$ with integer coefficients, $\mathbb{Z}[x]$. Then the set $S=\{1,x\}$ generates $\mathbb{Z}[x]$, since you can multiply $x$ by itself repeatedly to get any power $x^n$, then add copies of these along with copies of 1 to get any polynomial.

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    $\begingroup$ Dear Mike: It's a little hard for a newbie to see how you get elements of $\Bbb Z$, though. You might say that an empty product is $1$ by convention, and thus we get all elements of $\Bbb Z$, but it might be even easier to just say $\{1,x\}$ generates $\Bbb Z[x]$. Regards $\endgroup$ – rschwieb Sep 23 '14 at 13:21
  • $\begingroup$ That's a good point! It is now edited to me more clear. $\endgroup$ – Mike Earnest Sep 23 '14 at 15:18
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    $\begingroup$ slight correction, it should be $\pm$ all terms in the first equation. Taking additive inverses is usually allowed. $\endgroup$ – JHance Sep 23 '14 at 15:34
  • $\begingroup$ Good catch, also fixed. $\endgroup$ – Mike Earnest Sep 23 '14 at 15:37

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